0:07

activation energy and 'k'.

Â We are going to see this mathematically played out

Â in what is called the Arrhenius equation.

Â Our learning objective for this lesson is

Â to utilize the Arrhenius equation to determine

Â the rate constant,

Â the temperature, or the activation energy

Â of a reaction. We are going to see

Â the way that they interplay and relate off each other.

Â 0:32

This is the Arrhenius equation.

Â Lets look at the various component of this equation.

Â Talk about each one of them.

Â Many of them we have seen before. This is obviously the rate Talk about each one of them.

Â Many of them we have seen before. This is obviously the rate

Â constant 'k'.

Â We will be able to see how the rate constant

Â changes depending on the activation energy or the temperature.

Â We know that E sub A stand for the activation energy.

Â This is reported

Â and utilized either in kilojoules per mole

Â or joules per mole, we will see as we proceed on we have to make

Â sure our unit cancel properly.

Â 1:03

R is the ideal gas constant.

Â The ideal gas constant we are going to want to use

Â has units of joules

Â per mole.

Â Notice that I said kilojoules.

Â We cannot put these both in there with these same unit.

Â We either must have them both

Â in joules, so we could convert this one to joules

Â or we could have them both in kilojoules

Â and we convert this one to kilojoules, but those units are going to have to

Â cancel with each other

Â in order for this to work properly.

Â It is very common for students to mess that up.

Â They just plug numbers in and not thinking their units.

Â So I am talking about your units for a reason.

Â 1:42

OK if we are talking about units

Â what will the units of temperature have to be in?

Â As a hint look over at the gas constant

Â and what is the unit of temperature in it?

Â It is Kelvin, so you better put your temperature in Kelvin,

Â or this equation will not work.

Â So we have our 'k' or activation energy

Â our R and T and this E that is the exponent

Â the inverse of the natural log [ln]. our R and T and this E that is the exponent

Â the inverse of the natural log [ln].

Â 2:10

What about the A?

Â The A here is called the frequency factor.

Â It is a constant for the reaction

Â system. It takes into account

Â other avenues and things that are unique

Â to that reaction like

Â how do the molecule have to be oriented when they collide.

Â We will talk more about this frequency factor

Â later in our lesson.

Â 2:32

OK, We have this expression

Â but we want to do first is to think about

Â how this plays out with what we

Â already know is occurring on

Â the molecular level.

Â So over here we have got some conditions.

Â Lets think about changing activation energy.

Â As activation energy gets larger

Â and larger and we are look at this portion of the equation

Â we are raising it to a more and more negative value.

Â E to a larger and larger negative number.

Â How does that affect

Â the value of 'k'

Â If you are raising it to a large negative number

Â it becomes a smaller and smaller fraction.

Â So 'k' is going to go down.

Â Now does that make sense with what we know

Â about activation energy?

Â 3:22

Lets look and think about the mountain size.

Â If it is a really really high mountain

Â is it going to be a slower reaction?

Â There are going to be few molecules that can get over that is it going to be a slower reaction?

Â There are going to be few molecules that can get over that

Â so it is going to be a slower reaction

Â so you can expect a smaller 'k'.

Â 3:40

Lets think about temperature's role.

Â Now where is temperature in this equation?

Â Temperature is right here in the denominator.

Â As temperature goes up we are dividing by a larger

Â and larger number.

Â So it is going to decrease the

Â size of this portion of the number.

Â So it is to a smaller negative number.

Â If it is to a smaller negative number

Â then 'k' is going to become bigger and bigger.

Â Well, lets think about how that makes sense we what we know

Â about what is happen on the molecular level.

Â We have a certain mountain to climb.

Â As temperature is going up we have got more molecules We have a certain mountain to climb.

Â As temperature is going up we have got more molecules

Â that have enough energy to get over that mountain.

Â If there are more molecules that have enough energy to get over that mountain

Â wouldn't the reaction take place at a faster

Â speed? Therefore have a larger 'k' value.

Â 4:33

We are going to do some manipulations. Our goal

Â in the end is going to be is to have a

Â slope intercept form of a line.

Â So we will take the natural log of both sides.

Â Natural log of both sides

Â we are taking the natural log of a product here.

Â So we have the natural log of this product.

Â 5:14

Now I have not just rearranging but I have also done regrouping

Â and I want to look at what is happening.

Â I have moved this over here

Â to the left side, but I have also

Â grouped this together a little bit differently.

Â The reason I have done that is so it is in the

Â form y = mx +b.

Â So I have kind of lined it up underneath there so you know form y = mx +b.

Â So I have kind of lined it up underneath there so you know

Â what each thing is.

Â So what will it plot of the natural log of 'k'

Â That is my Y verses

Â 1 over t [1/T] that is my X.

Â It is definitely going to give you a straight line.

Â Well what is the slope going to be equal to?

Â 5:54

It is this whole portion here, that is m.

Â That is the slope, negative E sub A over R.

Â You could graphically determine

Â the activation energy by running this experiment

Â at various temperatures getting multiple k's

Â and obtaining an activation energy from the graph.

Â Here is an example. and obtaining an activation energy from the graph.

Â Here is an example.

Â 6:15

If you are give k's

Â and temperatures, you you need at first

Â take those k's and take the natural log of them.

Â You would take those temperatures and take

Â 1 over those temperatures.

Â Then plot out the natural log of k [ln k] along the y-axis.

Â 1 over t [1/t] along the x-axis.

Â Then you can take the best straight line through those points.

Â Once you have the best straight line through those points

Â you could take two values off of that line.

Â 6:50

You could do a change in Y

Â over a change of X and have a slope.

Â Once you have the slope, we know the slope is equal to

Â negative E sub A over R.

Â Then you could obtain the activation energy. negative E sub A over R.

Â Then you could obtain the activation energy.

Â 7:17

It you take this Arrhenius equation, and this is

Â the slope intercept form that equation

Â and you take it for two data points

Â let say for a K2

Â with a T2

Â and then do the same thing

Â a natural log of K1

Â with a 1 over T1 [1/T1]

Â take those two equations and subtract them

Â you can derivate a two point equation

Â for the Arrhenius equation.

Â And then you have got a relationship between

Â two temperatures and two rate constants

Â and it could be a way to obtain an activation energy.

Â But basically in this equation you have got

Â one, two, three, four, five variables

Â and if you know 4 of them you can solve

Â the fifth, and that is very often asked of you

Â within problems in this section.

Â So i have given you one here.

Â I have got a reaction in which I have given you the activation energy.

Â So this is E sub A.

Â I have told you the rate constant

Â k at 35 degrees Celsius .

Â So lets call this K1,

Â and lets call this T1.

Â And we want to know the K2

Â and we will call this T2.

Â Now remember you have got to have your

Â temperature in Kelvin.

Â So lets figure out T1 in Kelvin

Â it would be 273

Â plus 35.

Â And T2 would be

Â 273 + 0.

Â That is obviously 273 kelvin.

Â And lets get see that is 308.

Â 9:18

Just know that blob there

Â is a k.

Â So those are the temperatures

Â put everything we know into our Arrhenius equation.

Â I am going to call K2

Â what I don't know.

Â I like having it up on the top. It really

Â doesn't matter which you put on top and which you put on bottom

Â as long as your are consistent

Â that whatever k2 is you put it in the spot where

Â t2 needs to go.

Â K1 value is 1.35 x 10 ^ -4 seconds ^ -1.

Â 9:59

That is going to be equal to

Â E sub A over R.

Â Now here is where to have to watch the energy units.

Â I want to go ahead and convert my kilojoules per mole

Â to joules per mole.

Â That would be 1000 more.

Â So a 102 Thousand Joules

Â per mole is my activation energy.

Â My R is 8.314 Joules per mole times Kelvin

Â both of those in the denominator.

Â 10:57

Moles are going to cancel.

Â And now this Kelvin is in the denominator

Â of the denominator.

Â So the kelvins will cancel each other as well.

Â Because if you are in the denominator of the denominator thats

Â takes the reciprocal and puts it up in the numerator.

Â So my kelvins will cancel.

Â I am left with no units, and that is what you should have at this point.

Â OK so lets do a little work here.

Â When I combine, and you want to do all this.

Â Do the mathematics of this entire left side.

Â You will get a negative 5.107.

Â A very common mistake for students is they

Â lose this minus sign, don't do that.

Â That is going to be equation to the natural log of K2

Â over 1.35 x 10 ^ -4 seconds ^ -1.

Â Now I want to get into that K2 over 1.35 x 10 ^ -4 seconds ^ -1.

Â Now I want to get into that K2

Â how do I do that? I take E to both sides.

Â When I take E on this side

Â it cancels those out and I am left with K2

Â divided by 1.35 x 10 ^ -4 seconds ^ -1.

Â And when I take E to divided by 1.35 x 10 ^ -4 seconds ^ -1.

Â And when I take E to

Â negative 5.107

Â I have 0.00606.

Â 12:40

That will give me a K2 value

Â of 8.18 x 10 ^-7 second ^ -1. That will give me a K2 value

Â of 8.18 x 10 ^-7 second ^ -1.

Â Lets think if this makes sense.

Â What did we do to our temperature?

Â We went from 35 where we knew a K

Â down to zero.

Â So we dropped the temperature.

Â If we drop the temperature, the rate should decrease,

Â and it does so by having a smaller k.

Â if our k smaller? Well certainly it is.

Â So at least we know that this number makes sense for the work we have done.

Â So with out Arrhenius equation we see

Â the relationship between temperature

Â activation energy and k.

Â We need to mention a little bit about that frequency factor A that

Â comes out of that Arrhenius equation.

Â 13:34

That quantity frequency factor is going to be

Â dependent on a couple of things.

Â Number one it takes into account

Â the collision frequency.

Â Therefore it is called the frequency factor.

Â With that, you take into account the state of matter, for example.

Â Something that is in a gas state With that, you take into account the state of matter, for example.

Â Something that is in a gas state

Â free to move and they move very very rapidly

Â so they can collide with each other

Â but there is great distances between them.

Â But if they are in a liquid state

Â they still can move pretty quickly but they

Â much much closer together.

Â They will collide more frequently

Â so that takes into account the state.

Â If it is a solid, they do not move very freely

Â and that cannot collide very frequently. So that would

Â have a different frequency factor.

Â It also takes into account.

Â orientation of the molecules

Â because sometimes one collision even

Â if it has a lot of energy won't result in a reaction

Â and we see that here in the top

Â image. In the top image if they were to collide

Â no reaction would occur.

Â But if they collided and the green ran into the green

Â then there could be a reaction.

Â If there is an orientation necessary

Â to have a collision

Â that will come out in that Arrhenius equation.

Â In conclusion with lesson and out learning objective here

Â we see that there is mathematical relationship between temperature

Â activation energy and the rate constant k.

Â We saw how to graph it out

Â in order to obtain a linear relationship.

Â How to plug into a two point value equation

Â with a K1, T1

Â K2, T2

Â how to be able to obtain an unknown out of that equation.

Â Lastly we see how the frequency

Â factor A plays a role in rates of reactions. Lastly we see how the frequency

Â factor A plays a role in rates of reactions.

Â